For the reaction between $M n O_{4}^{-}$ and $C_{2} O_{4}^{-}$ in basic solution, the unbalanced equation is as follows:
$M n O_{4}^{-} + C_{2} O_{4}^{2 -} ⟶ M n O_{2} (s) + C O_{3}^{2 -}$
In a balanced equation, the number of ${O H}^{-}$ ions on the right is :

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Solution

The balanced equation is as follows:
$4 M n O_{4}^{-} + 3 C_{2} O_{4}^{2 -} + 2 H_{2} O \to 4 M n O_{2} + 6 C O_{3}^{2 -} + 4 O H^{-}$
Hence, the number of $O H^{-}$ on the right is $4$ .

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